The general solution of the differential equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the differential equation: $(1+y^2) dx = (	an^{-1} y - x) dy$ Rearranging the terms,we get: $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2}$

Vedclass · Accepted Answer

$x e^{	an^{-1} y} = e^{	an^{-1} y} (	an^{-1} y - 1) + c$

Vedclass · Answer

(D) Given the differential equation: $(1+y^2) dx = (	an^{-1} y - x) dy$ Rearranging the terms,we get: $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2}$ $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{	an^{-1} y}{1+y^2}$ This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{1}{1+y^2}$ and $Q = \frac{	an^{-1} y}{1+y^2}$. The integrating factor $(IF)$ is given by $IF = e^{\int P dy} = e^{\int \frac{1}{1+y^2} dy} = e^{	an^{-1} y}$. The general solution is $x \cdot IF = \int (Q \cdot IF) dy + c$. $x e^{	an^{-1} y} = \int \frac{	an^{-1} y}{1+y^2} e^{	an^{-1} y} dy + c$. Let $t = 	an^{-1} y$,then $dt = \frac{1}{1+y^2} dy$. $x e^{	an^{-1} y} = \int t e^t dt + c$. Using integration by parts: $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t(t-1)$. Substituting back $t = 	an^{-1} y$,we get: $x e^{	an^{-1} y} = e^{	an^{-1} y} (	an^{-1} y - 1) + c$.

The general solution of the differential equation $(1+y^2) dx = ( an^{-1} y - x) dy$ is

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $(x \log x) \frac{dy}{dx} + y = 2x \log x$ for $x \geq 1$. Then $y(e)$ is equal to:

If $y(t)$ is a solution of $(1 + t)\frac{dy}{dt} - ty = 1$ and $y(0) = -1$,then $y(1)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $x^3 dy + (xy - 1) dx = 0, x > 0$,with $y(\frac{1}{2}) = 3 - e$. Then $y(1)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $x \log _e x \frac{d y}{d x}+y=x^2 \log _e x, (x > 1)$. If $y(2)=2$,then $y(e)$ is equal to

The differential equation $\frac{dy}{dx} = \frac{x+y}{1+x^2}$ is a . . . . . . differential equation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module